3.340 \(\int \frac{x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx\)

Optimal. Leaf size=156 \[ \frac{b x^{m+1} (b c (1-m)-a d (3-m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^2}+\frac{d^2 x^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c (m+1) (b c-a d)^2}+\frac{b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)} \]

[Out]

(b*x^(1 + m))/(2*a*(b*c - a*d)*(a + b*x^2)) + (b*(b*c*(1 - m) - a*d*(3 - m))*x^(
1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(2*a^2*(b*c - a
*d)^2*(1 + m)) + (d^2*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*
x^2)/c)])/(c*(b*c - a*d)^2*(1 + m))

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Rubi [A]  time = 0.468687, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136 \[ -\frac{b x^{m+1} (a d (3-m)-b (c-c m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^2}+\frac{d^2 x^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c (m+1) (b c-a d)^2}+\frac{b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]  Int[x^m/((a + b*x^2)^2*(c + d*x^2)),x]

[Out]

(b*x^(1 + m))/(2*a*(b*c - a*d)*(a + b*x^2)) - (b*(a*d*(3 - m) - b*(c - c*m))*x^(
1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(2*a^2*(b*c - a
*d)^2*(1 + m)) + (d^2*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*
x^2)/c)])/(c*(b*c - a*d)^2*(1 + m))

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Rubi in Sympy [A]  time = 95.4734, size = 128, normalized size = 0.82 \[ \frac{d^{2} x^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{d x^{2}}{c}} \right )}}{c \left (m + 1\right ) \left (a d - b c\right )^{2}} - \frac{b x^{m + 1}}{2 a \left (a + b x^{2}\right ) \left (a d - b c\right )} - \frac{b x^{m + 1} \left (- a d m + 3 a d + b c m - b c\right ){{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{2 a^{2} \left (m + 1\right ) \left (a d - b c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**m/(b*x**2+a)**2/(d*x**2+c),x)

[Out]

d**2*x**(m + 1)*hyper((1, m/2 + 1/2), (m/2 + 3/2,), -d*x**2/c)/(c*(m + 1)*(a*d -
b*c)**2) - b*x**(m + 1)/(2*a*(a + b*x**2)*(a*d - b*c)) - b*x**(m + 1)*(-a*d*m +
3*a*d + b*c*m - b*c)*hyper((1, m/2 + 1/2), (m/2 + 3/2,), -b*x**2/a)/(2*a**2*(m
+ 1)*(a*d - b*c)**2)

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Mathematica [A]  time = 0.112833, size = 127, normalized size = 0.81 \[ \frac{x^{m+1} \left (a^2 d^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )-a b c d \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )+b c (b c-a d) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )\right )}{a^2 c (m+1) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]  Integrate[x^m/((a + b*x^2)^2*(c + d*x^2)),x]

[Out]

(x^(1 + m)*(-(a*b*c*d*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])
+ a^2*d^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] + b*c*(b*c -
a*d)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)]))/(a^2*c*(b*c - a*
d)^2*(1 + m))

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Maple [F]  time = 0., size = 0, normalized size = 0. \[ \int{\frac{{x}^{m}}{ \left ( b{x}^{2}+a \right ) ^{2} \left ( d{x}^{2}+c \right ) }}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^m/(b*x^2+a)^2/(d*x^2+c),x)

[Out]

int(x^m/(b*x^2+a)^2/(d*x^2+c),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x^{m}}{{\left (b x^{2} + a\right )}^{2}{\left (d x^{2} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^m/((b*x^2 + a)^2*(d*x^2 + c)),x, algorithm="maxima")

[Out]

integrate(x^m/((b*x^2 + a)^2*(d*x^2 + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{x^{m}}{b^{2} d x^{6} +{\left (b^{2} c + 2 \, a b d\right )} x^{4} + a^{2} c +{\left (2 \, a b c + a^{2} d\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^m/((b*x^2 + a)^2*(d*x^2 + c)),x, algorithm="fricas")

[Out]

integral(x^m/(b^2*d*x^6 + (b^2*c + 2*a*b*d)*x^4 + a^2*c + (2*a*b*c + a^2*d)*x^2)
, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**m/(b*x**2+a)**2/(d*x**2+c),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x^{m}}{{\left (b x^{2} + a\right )}^{2}{\left (d x^{2} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^m/((b*x^2 + a)^2*(d*x^2 + c)),x, algorithm="giac")

[Out]

integrate(x^m/((b*x^2 + a)^2*(d*x^2 + c)), x)